Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int MAXN = 100005;
using namespace std;
int n, m, q, v, u, a;
vector<int> E[MAXN], c[MAXN];
int L[MAXN], P[MAXN][17], M[MAXN][17][10], res[20], aux[20], cc[10];
void dfs(int cur, int prev) {
  P[cur][0] = prev;
  L[cur] = L[prev] + 1;
  for (auto to : E[cur])
    if (to != prev) dfs(to, cur);
}
void precalc() {
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i <= n; i++) P[i][j] = P[P[i][j - 1]][j - 1];
  for (int j = 1; (1 << j) <= n; j++) {
    for (int i = 1; i <= n; i++) {
      fill(aux, aux + 20, 1000000);
      merge(M[i][j - 1], M[i][j - 1] + 10, M[P[i][j - 1]][j - 1],
            M[P[i][j - 1]][j - 1] + 10, aux);
      for (int h = 0; h < 10; h++) M[i][j][h] = aux[h];
    }
  }
}
int LCA(int x, int y) {
  if (L[x] < L[y]) swap(x, y);
  for (int j = 16; j >= 0; j--)
    if (L[x] - (1 << j) >= L[y]) {
      merge(res, res + 10, M[x][j], M[x][j] + 10, aux);
      for (int h = 0; h < 10; h++) res[h] = aux[h];
      x = P[x][j];
    }
  if (x == y) return x;
  for (int j = 16; j >= 0; j--)
    if (P[x][j] != 0 && P[x][j] != P[y][j]) {
      merge(res, res + 10, M[x][j], M[x][j] + 10, aux);
      for (int h = 0; h < 10; h++) res[h] = aux[h];
      merge(res, res + 10, M[y][j], M[y][j] + 10, aux);
      for (int h = 0; h < 10; h++) res[h] = aux[h];
      x = P[x][j];
      y = P[y][j];
    }
  merge(res, res + 10, M[x][0], M[x][0] + 10, aux);
  for (int h = 0; h < 10; h++) res[h] = aux[h];
  merge(res, res + 10, M[y][0], M[y][0] + 10, aux);
  for (int h = 0; h < 10; h++) res[h] = aux[h];
  return P[x][0];
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &v, &u);
    E[v].emplace_back(u);
    E[u].emplace_back(v);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a);
    c[a].emplace_back(i);
  }
  dfs(1, 0);
  for (int i = 0; i <= n; i++)
    for (int j = 0; (1 << j) <= n; j++)
      for (int h = 0; h < 10; h++) M[i][j][h] = 1000000;
  for (int i = 1; i <= n; i++) sort((c[i]).begin(), (c[i]).end());
  for (int i = 1; i <= n; i++)
    for (int h = 0; h < min((int)c[i].size(), 10); h++) M[i][0][h] = c[i][h];
  precalc();
  while (q--) {
    scanf("%d%d%d", &v, &u, &a);
    fill(cc, cc + 20, 1000000);
    fill(res, res + 20, 1000000);
    int lca = LCA(v, u), sz = 0;
    for (int h = 0; h < min((int)c[lca].size(), 10); h++) cc[h] = c[lca][h];
    merge(res, res + 10, cc, cc + 10, aux);
    for (int h = 0; h < min(a, 10) && aux[h] != 1000000; h++) sz++;
    printf("%d ", sz);
    for (int h = 0; h < sz; h++) printf("%d ", aux[h]);
    printf("\n");
  }
}